Ring: Proof of Group

1. The problem statement, all variables and given/known data
Let R be the ring of all 2*2 matrices, over Zp, p a prime. Let G be the set of elements x in the ring R such that det x ≠ 0. Prove that G is a group.

2. Relevant equations
Matrix is invertible in ring R.

3. The attempt at a solution
Group properties and ring properties are similar I think.
Group and Ring - closure, associativity, identity (zero in Rings?)
Is this how I am supposed to approach the problem? By proving the common properties of a group and ring?

Group properties and ring properties are similar I think.
Group and Ring - closure, associativity, identity (zero in Rings?)
Is this how I am supposed to approach the problem? By proving the common properties of a group and ring?

##G## will be a group under multiplication, not addition. So you can ignore the addition and address the following questions:

1. What is the multiplicative identity? Is it contained in ##G##?
2. Is matrix multiplication associative?
3. Is the set of matrices with nonzero determinant closed under multiplication and inverses?

Hint for part 3: can you prove and apply the identities ##\det(AB) = \det(A)\det(B)## and ##\det(I) = 1##?

Staff: Mentor

1. The problem statement, all variables and given/known data
Let R be the ring of all 2*2 matrices, over Zp, p a prime. Let G be the set of elements x in the ring R such that det x ≠ 0. Prove that G is a group.

2. Relevant equations
Matrix is invertible in ring R.

3. The attempt at a solution
Group properties and ring properties are similar I think.
Group and Ring - closure, associativity, identity (zero in Rings?)
Is this how I am supposed to approach the problem? By proving the common properties of a group and ring?

The elements of G already belong to a ring, so these elements already satisfy all of the properties of a ring (which must include an addition operation and a multiplication operation). If you list the properties that a group must satisfy, this should be pretty easy to prove.

##G## will be a group under multiplication, not addition. So you can ignore the addition and address the following questions:

1. What is the multiplicative identity? Is it contained in ##G##?
2. Is matrix multiplication associative?
3. Is the set of matrices with nonzero determinant closed under multiplication and inverses?

Hint for part 3: can you prove and apply the identities ##\det(AB) = \det(A)\det(B)## and ##\det(I) = 1##?

1. Multiplicative identity in a ring is 1, right? But 1 is not in G, because G consists of the set of matrices for which ad - bc ≠ 0, but (1) is not a 2*2 matrix.
2. Yes; [(ab)c]ij = (ab)ik ckj = (ailblk)ckj = ail(blkckj) = ail(bc)lj = [a(bc)]ij
3. Yes, for A= (a b c d), and B= (e f g h), it was proven that det(A)det(B)=det(AB), and det(I)= 1 by 1/ad-bc (ad-bc 0 0 ad-bc) since 1/ad-bc where ad-bc≠0 is the inverse of matrix M. The set is closed.

1. Multiplicative identity in a ring is 1, right? But 1 is not in G, because G consists of the set of matrices for which ad - bc ≠ 0, but (1) is not a 2*2 matrix.

We often use the notation "1" for the multiplicative identity in a ring, but it doesn't have to literally mean the number 1. In this case, the ring consists of 2x2 matrices over ##Z_p##, so if there is an identity, it must be a 2x2 matrix over ##Z_p##. What is the definition of "identity"?

3. Yes, for A= (a b c d), and B= (e f g h), it was proven that det(A)det(B)=det(AB), and det(I)= 1 by 1/ad-bc (ad-bc 0 0 ad-bc) since 1/ad-bc where ad-bc≠0 is the inverse of matrix M. The set is closed.

Where did you use the fact that the matrix elements are from ##Z_p## (where ##p## is prime), as opposed to ##Z_n## for some nonprime integer? Also, can you explain in more detail how you concluded that the set is closed? How do the determinant formulas prove this?

We often use the notation "1" for the multiplicative identity in a ring, but it doesn't have to literally mean the number 1. In this case, the ring consists of 2x2 matrices over ##Z_p##, so if there is an identity, it must be a 2x2 matrix over ##Z_p##. What is the definition of "identity"?

Where did you use the fact that the matrix elements are from ##Z_p## (where ##p## is prime), as opposed to ##Z_n## for some nonprime integer? Also, can you explain in more detail how you concluded that the set is closed? How do the determinant formulas prove this?

I failed to mention that the elements are from Zp, p being a prime. Zp is a field if and only if p is prime, and the properties hold true because of this.
Since the determinant is nonzero, and the inverse has been solved for: Det(M)* I = 1, the set is closed under multiplication and inverses. A group must contain an element and its inverse, which this set of matrices does.

1. How do you know that ##M## has an inverse?
2. How do you know that ##\det(M^{-1}) \neq 0##?

M has an inverse because M= (a b c d) has another matrix such that M * (other matrix) = Identity Matrix. We know that the det(M-1) ≠ 0 because det(M) ≠0, and 1/det(M)≠0, ever. This ring is specifically an integral domain, I believe.

M has an inverse because M= (a b c d) has another matrix such that M * (other matrix) = Identity Matrix.

How do you know this is true? It would not be true if the matrix elements were from ##Z_n## where ##n## is not prime.

We know that the det(M-1) ≠ 0 because det(M) ≠0, and 1/det(M)≠0, ever.

OK, so you're using ##\det(M)\det(M^{-1}) = \det(M M^{-1}) = \det(I) = 1##. You need to state this more explicitly.

This ring is specifically an integral domain, I believe.

An integral domain is commutative by definition. Matrix multiplication is not commutative.

The property of a ring is closure under multiplication and identity exists.
So since the ring is a subgroup of the M(2*2) then it is a group itself.

What ring is a subgroup of ##M_{2\times 2}##? I'm not following your logic. You need to explain why ##G## is closed under multiplication. If ##A## and ##B## are two matrices with nonzero determinant, what can you say about ##\det(AB)##?

What ring is a subgroup of ##M_{2\times 2}##? I'm not following your logic. You need to explain why ##G## is closed under multiplication. If ##A## and ##B## are two matrices with nonzero determinant, what can you say about ##\det(AB)##?

...because ##\det(AB) = \det(A)\det(B)##. And how do you know that if ##\det(A)## and ##\det(B)## are nonzero, then so is ##\det(A)\det(B)##? Keep in mind that we are working in ##Z_p##, not the real numbers, so you have to justify what may seem like standard facts. Again, this would not be true in ##Z_n## for nonprime ##n##.

...because ##\det(AB) = \det(A)\det(B)##. And how do you know that if ##\det(A)## and ##\det(B)## are nonzero, then so is ##\det(A)\det(B)##? Keep in mind that we are working in ##Z_p##, not the real numbers, so you have to justify what may seem like standard facts. Again, this would not be true in ##Z_n## for nonprime ##n##.

No, it's true for every matrix ring that [itex]\det(AB) = \det(A)\det(B)[/itex]. The question here is: if [itex]\det(A)[/itex] and [itex]\det(B)[/itex] are both non-zero, is it necessarily the case that [itex]\det(A)\det(B)[/itex] is non-zero?

This is equivalent to showing that, for [itex]a \in \mathbb{Z}_p[/itex] and [itex]b \in \mathbb{Z}_p[/itex], if [itex]a \neq 0[/itex] and [itex]b \neq 0[/itex] then [itex]ab \neq 0[/itex].

This isn't true for general [itex]\mathbb{Z}_n[/itex]: for example in [itex]\mathbb{Z}_{10}[/itex] we have [itex]2 \times 5 = 0[/itex]. So what's special about [itex]\mathbb{Z}_p[/itex] for prime [itex]p[/itex]?